a rock is dropped from a 20.0m bridge falls into the river below.
a.)at whqat velocity does it strike the water?
b.)how much time did the rock take to reach the water?
To find the answers, we can use the equations of motion and the laws of gravity. Let's go step by step:
a.) To determine the velocity at which the rock strikes the water (final velocity), we can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity (unknown)
u = initial velocity (0 m/s as the rock is dropped)
a = acceleration due to gravity (approximately 9.8 m/s^2)
s = displacement (20.0 m, negative since the rock is falling downwards)
Rearranging the equation, we have:
v^2 = 0^2 + 2 * 9.8 * -20.0
v^2 = 0 + (-392)
v^2 = -392
Taking the square root of both sides, we find:
v = √(-392)
It is important to note that the square root of a negative number is not a real number. Therefore, the rock does not have a velocity with which it strikes the water. Instead, it approaches the water with increasing speed but no definite velocity.
b.) To calculate the time taken for the rock to reach the water, we can use the equation:
s = ut + (1/2)at^2
where:
s = displacement (20.0 m, negative as the rock is falling downwards)
u = initial velocity (0 m/s)
a = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (unknown)
Rearranging the equation, we have:
20.0 = 0 * t + (1/2) * 9.8 * t^2
20.0 = 4.9t^2
Dividing both sides by 4.9, we find:
4.08 = t^2
Taking the square root of both sides, we get:
t ≈ √4.08
t ≈ 2.02 seconds
Therefore, it takes approximately 2.02 seconds for the rock to reach the water from the 20.0m bridge.